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Homework Help: Second derivative=0; stable/unstable equilibrium?

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Describe how to determine whether an equilibrium is stable or unstable when [d2U/dx2]_0 = 0

    From Classical Dynamics - Ch 2 #45 - Marion Thornton

    2. Relevant equations AND 3. The attempt at a solution

    When second derivative positive, equilibrium is stable. When second derivative negative, equilibrium unstable.

    When second derivative is 0, you have to do another test. I was under the impression that you could find a third derivative, fourth derivative, until you get a positive/negative value for x=0 and use the results to get stability, with the same rule as above? And if second and subsequent derivatives are 0, it's a neutral equilibrium.

    The main reason I'm confused is the book had a weird solution that didn't make sense to me. They expanded the potential about the equilibrium point (how? why?), found the first derivative, and said that it's stable when n is even and unstable when n is odd. But that doesn't make sense to me. How could they make a conclusion like that if they don't know what U is?

    Maybe I'm misunderstanding the question's notation - I understood it to mean "second derivative at x=0 is 0."
  2. jcsd
  3. Oct 4, 2009 #2


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    What is n? Is your potential of the form U = kxn (n > 0)? If so, sketch the potential for even and odd values of n and see for yourself what happens at x = 0.
  4. Oct 5, 2009 #3


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    Ermmm...what the heck is a "neutral equilibrium"?:confused:

    I'm guessing that the expansion is a Taylor expansion...you know how to do those right? What is the derivative of potential equal to physically? What should that quantity be near a stable/unstable equilibrium point?
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